FLUCTUATING STRESSES SUBJECT: Design of machine elements Submitted to- Dhaval Darji Sir Mechanical Department By- 130010119100 130010119101 130010119102 130010119103 130010119104

Fatigue load The loads, which vary in magnitude and/or direction with respect to time, are known as fatigue, fluctuating or alternating loads. It has been observed that, when the mechanical component is subjected to fluctuating loads, it fails at a stress considerably below the ultimate strength and quite frequently even below the yield strength. Such failure is Fatigue failure.

B Instantaneuos Fast Fracture! A Crack nucleation and Growth

Fluctuating stresses When the mechanical component is subjected to the fatigue or fluctuating load, the stress induced is known as fluctuating stress.

Repeated & Reversed Stress an element subjected to a repeated and alternating tensile and compressive stresses. Continuous total load reversal over time

Definitions: = Alternating stress = Mean stress = R value: R = 0, repeated and one direction, i.e. stress cycles from 0 to max value. R =-1, Fully reversed (R-R Moore)

1.Repeated and Reversed Stress The average or mean stress is zero.

Fluctuating Stress When an element experiences alternating stress, but the mean stress is NOT zero. Load varies between P and Q over time

2.Fluctuating Stress Example Bending of Rocker Arm Valve Spring Force Valve Open Valve Closed Tension in Valve Stem Valve Closed Valve Spring Force Valve Open

Tensile Stress w/ Tensile Mean

Partially Reversed w/ Tensile Mean smax is tensile and smin is compressive

Partially Reversed w/ Compressive Mean smax is tensile and smin is compressive

Compressive Stress w/ Compressive Mean smax and smin are both compressive

Repeated – One Direction Stress

Fatigue Failure, S-N Curve Test specimen geometry for R.R. Moore rotating beam machine. The surface is polished in the axial direction. A constant bending load is applied. Motor Load Rotating beam machine – applies fully reverse bending stress Typical testing apparatus, pure bending

Fatigue Failure, S-N Curve Finite life Infinite life S′e = endurance limit of the specimen Se ′

Design for Finite Life Sn = a (N)b equation of the fatigue line Se 106 103 A B N S Sf 5x108 103 A B Point A Sn = .9Sut N = 103 Point A Sn = .9Sut N = 103 Point B Sn = Se N = 106 Point B Sn = Sf N = 5x108

Sn = a (N)b Sn Se ( ) ⅓ Sn n a log .9Sut = log a + b log 103 log Sn = log a + b log N a = (.9Sut)2 Se b .9Sut 1 3 log log .9Sut = log a + b log 103 log Se = log a + b log 106 Sn = Se ( N 106 ) ⅓ .9Sut log Sn Kf a = n Design equation Calculate Sn and replace Se in the design equation

Design of components subjected to fluctuating stresses for infinite life Mean stress Alternating stress m a Sy Yield line Gerber curve Se Sy Soderberg line Sut Goodman line

The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram Alternating stress m a Sy Yield line Se Safe zone C Goodman line Sy Sut Mean stress

a Sy Yield line - Syc Se Safe zone Sut Goodman line Safe zone - m Sy

a a m a m a = a + m = a + m = m > 0 m ≤ 0 nf Se Sut Sn Fatigue, m ≤ 0 Fatigue, m > 0 a nf Se 1 = Sut a m + Infinite life a = Se nf Finite life Sn 1 = Sut a m + a + m = Sy ny Yield Se a + m = Sy ny Yield C Safe zone Safe zone - m - Syc Sy Sut +m

Combined loading xya alternating component of shear stress All four components of stress exist, xa alternating component of normal stress xm mean component of normal stress xya alternating component of shear stress xym mean component of shear stress Calculate the alternating and mean principal stresses, 1a, 2a = (xa /2) ± (xa /2)2 + (xya)2 1m, 2m = (xm /2) ± (xm /2)2 + (xym)2

a′ = (1a + 2a - 1a2a)1/2 m′ = (1m + 2m - 1m2m)1/2 ′a ′m nf Calculate the alternating and mean von Mises stresses, a′ = (1a + 2a - 1a2a)1/2 2 m′ = (1m + 2m - 1m2m)1/2 2 Fatigue design equation nf Se 1 = Sut ′a ′m + Infinite life

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